A Fairly Simple Probability Question

[David Sklansky]

I'm curious as to whether others do this the same way I do.

You are dealt two cards, both of which contain a randomly chosen real number between zero and one. What is the probability that they add up to more than 1.6?

[valenzuela]

easy, ill put my calculations here.( it wont make too much sense)

40% of the time we are still alive after the first card.
When its 0,6 we have a 0% chance of winning
when it 1 we have a 40% chance of winning, on average we will have a 20% chane of winning.
There by 0,4 x 0,2 = 0,08

8%.

[surftheiop]

hmm is it bad to compare infinities to each other?

[Jcrew]

I default to calculus/conditional probability but its probably easier to map them along the y and x axis from [0,1], and take the area of the unit square above the y = -x + 1.6 line.

[bunny]

I use the area method, but I'm a geometer so I always like pictures. /images/graemlins/smile.gif

[jogsxyz]

It's solvable. 2/5 times 2/5 times 1/2 = 2/25

Showing the proof is tougher.

[goodsamaritan]

x + y = 1.6
y = 1.6 - x

We want the area above the curve for when 0<x<1 and 0<y<1.

That area is a triangle with a base of .4 and height of .4. Area = .5(.4*.4) = .08

[jogsxyz]

I drew a unit square. Across is the x-axis and up and down is the y-axis. Plotted the points which would sum 1.6 of greater. They were all within a triangle on the top right hand corner. Then calculate the area of the triangle.

2/5 X 2/5 X 1/2 = 2/25

[Skidoo]

I'm gonna crap-shoot this one.

For some reason, the fact that the cards can take any real value makes me think of just adding them straight. That's my guess, so the answer would be the same as for the following:

If you have a 20 ounce beer mug filled to a random fraction of its capacity, what is the probability it will have more than 16 ounces of beer in it?

That is, 1 in 5.

[madnak]

I do it like everyone else, draw the line and then "box" the 0-1.0 and check the area of the triangle.

[bigpooch]

I just picture the unit square and take the area of the
appropriate triangle which is just half the area of the
square which includes (1,1) and (0.6, 0.6) as vertices.

[thylacine]

Integrate the probability distribution p(x,y) over the triangle determined by

x<=1
y<=1
x+y>1.6

If p(x,y)=1 for all x,y, then the answer is just the area of the triangle, namely 0.08.

Why do you ask?

[Phil153]

Each card must be between 0.6 and 1. That's 0.4*0.4 = 16% of all two card pairs. Half of these combinations will have an average above 0.8.

If put on the spot I'd probably do it Valen's way.

[Borodog]

y+x >= 1.6, y >= 1.6-x. Hence the area to the upper right of the line y=1.6-x, to the left of the line x=1, and below the line y=1 satisfies the equation. This area divided by the area bounded by x=0,1 and y=0,1 (i.e. 1) is the probability you want. It is (1/2)(0.4)^2 = 0.08.

[valenzuela]

bah equations are boring, its more fun my way.

[PairTheBoard]

Pretend there are 5 cards. There are 25 combinations. A (4,5) is a good one as well as a (5,4). That's 2/25 = .08

PairTheBoard

[madnak]

What about (5,5) and (4,4)? And isn't using discrete measures imprecise at best in this kind of situation?

[PairTheBoard]

[ QUOTE ]
What about (5,5) and (4,4)? And isn't using discrete measures imprecise at best in this kind of situation?

[/ QUOTE ]

(4,4) would represent two numbers below 0.8 so wouldn't work. (5,5) has to be thrown out to get the right answer.

PairTheBoard

[Borodog]

[ QUOTE ]
[ QUOTE ]
What about (5,5) and (4,4)? And isn't using discrete measures imprecise at best in this kind of situation?

[/ QUOTE ]

(4,4) would represent two numbers below 0.8 so wouldn't work. (5,5) has to be thrown out to get the right answer.

PairTheBoard

[/ QUOTE ]

Lol.

[Skidoo]

[ QUOTE ]
Pretend there are 5 cards. There are 25 combinations. A (4,5) is a good one as well as a (5,4). That's 2/25 = .08

PairTheBoard

[/ QUOTE ]

I think I see what you did, and it's an interesting analogy to the method of areas.

The range of acceptable values for either card is 0.6 to 1. For each value x in that range for the first card, the average corresponding acceptable value for the second card is

1 - (x - 0.6) / 2.

So, since the average acceptable value for the first card is 0.8, the average corresponding acceptable value for the second card is

1 - (0.8 - 0.6) / 2 = 1 - 0.1 = 0.9.

Since P(x) = 1 - x, the required minimum sum outcome probabilities divided by total outcome probabilities is

(1 - 0.8) * (1 - 0.9) / (1 - .5)^2 = 0.2 * 0.1 / 0.25 = 0.08,

the correct answer.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Pretend there are 5 cards. There are 25 combinations. A (4,5) is a good one as well as a (5,4). That's 2/25 = .08

PairTheBoard

[/ QUOTE ]

I think I see what you did, and it's an interesting analogy to the method of areas.

The range of acceptable values for either card is 0.6 to 1. For each value x in that range for the first card, the average corresponding acceptable value for the second card is

1 - (x - 0.6) / 2.

So, since the average acceptable value for the first card is 0.8, the average corresponding acceptable value for the second card is

1 - (0.8 - 0.6) / 2 = 1 - 0.1 = 0.9.

Since P(x) = 1 - x, the required minimum sum outcome probabilities divided by total outcome probabilities is

(1 - 0.8) * (1 - 0.9) / (1 - .5)^2 = 0.2 * 0.1 / 0.25 = 0.08,

the correct answer.

[/ QUOTE ]

That must have been what I was thinking.

PairTheBoard

[arahant]

[ QUOTE ]
easy, ill put my calculations here.( it wont make too much sense)

40% of the time we are still alive after the first card.
When its 0,6 we have a 0% chance of winning
when it 1 we have a 40% chance of winning, on average we will have a 20% chane of winning.
There by 0,4 x 0,2 = 0,08

8%.

[/ QUOTE ]

Yeah, that's how i do it...what's up with all these integrals and area calculations and [censored]...have some intuition, people! /images/graemlins/smile.gif

[Skidoo]

In general, p(x) = (2 - x)^2 / 2, where x is the sum of the two card values and greater than 1.

[madnak]

[ QUOTE ]
[ QUOTE ]
easy, ill put my calculations here.( it wont make too much sense)

40% of the time we are still alive after the first card.
When its 0,6 we have a 0% chance of winning
when it 1 we have a 40% chance of winning, on average we will have a 20% chane of winning.
There by 0,4 x 0,2 = 0,08

8%.

[/ QUOTE ]

Yeah, that's how i do it...what's up with all these integrals and area calculations and [censored]...have some intuition, people! /images/graemlins/smile.gif

[/ QUOTE ]

You really don't need the integral for the geometric solution. Integrals are for curves, not triangles. Still, I prefer the algebraic way. But I would reach the solution geometrically - what can I say?

[valenzuela]

bump, I want to know how did the OP solved it.

[holmansf]

[ QUOTE ]
have some intuition, people!

[/ QUOTE ]

The geometric method is the most intuitive for me.

[lastcardcharlie]

[ QUOTE ]

You are dealt two cards, both of which contain a randomly chosen real number between zero and one.


[/ QUOTE ]

Bearing in mind that this can never actually be done, i.e. that there can never be a card-dealing machine that deals with the reals (or the integers for that matter) in their totality, the question is not as simple or meaningful as it might seem.

Classically it would treated as some sort of "limiting case" of things that can in reality be done. Suppose first that the cards contain numbers between 0 and 10 and are required to add up to more than 16. There are 1 + 2 + 3 + 4 combinations that will give this, among a total number of combinations of 11!/2!9!.

Now consider that case as the case when n = 1, and then the general case becomes a selection of cards containing numbers between 0 and 10n which are required to add up to more than 16n. There are

1 + 2 + .. + 4n = 4n(4n + 1)/2

combinations that will give this, among a total number of combinations of

(10n + 1)!/2!(10n - 1)! = 10n(10n + 1)/2

The probability of success is therefore

4n(4n + 1)/10n(10n + 1).

As n tends to "infinity" (which will still probably only give the case for the rationals between 0 and 1, and some sort of nightmare completion process will be required to obtain the reals), the probability of success will tend to

4/10 * 4/10 = 0.16

While everyone is saying 0.08. Goddam, I must have calculated wrong somewhere.

[goodsamaritan]

[ QUOTE ]
[ QUOTE ]

You are dealt two cards, both of which contain a randomly chosen real number between zero and one.


[/ QUOTE ]

Bearing in mind that this can never actually be done, i.e. that there can never be a card-dealing machine that deals with the reals (or the integers for that matter) in their totality, the question is not as simple or meaningful as it might seem.

Classically it would treated as some sort of "limiting case" of things that can in reality be done. Suppose first that the cards contain numbers between 0 and 10 and are required to add up to more than 16. There are 1 + 2 + 3 + 4 combinations that will give this, among a total number of combinations of 11!/2!9!.

Now consider that case as the case when n = 1, and then the general case becomes a selection of cards containing numbers between 0 and 10n which are required to add up to more than 16n. There are

1 + 2 + .. + 4n = 4n(4n + 1)/2

combinations that will give this, among a total number of combinations of

(10n + 1)!/2!(10n - 1)! = 10n(10n + 1)/2

The probability of success is therefore

4n(4n + 1)/10n(10n + 1).

As n tends to "infinity" (which will still probably only give the case for the rationals between 0 and 1, and some sort of nightmare completion process will be required to obtain the reals), the probability of success will tend to

4/10 * 4/10 = 0.16

While everyone is saying 0.08. Goddam, I must have calculated wrong somewhere.

[/ QUOTE ]

What crack did you smoke when you wrote that?

[nlc315]

I haven't taken a stat class yet...I'm finishing up the calculus this semester, but I need to learn this type of thing for my major. If I had to render a guess:

Given: 2 cards dealt, with each card having 2 possible numbers on them.

Probability of being dealt 0,0: 25%
Probability of being dealt 0,1 or 1,0: 50%
Probability of being dealt 1,1: 25%

The only possible outcome where the sum of the numbers on both cards > 1.6 is with 1+1=2.

So...25%?

[nlc315]

oh...I misread the question ;-) I wondered if there was some trick or something.

There are an infinite amount of real numbers between 0 and 1. So the probability of being dealt a hand at all would be essentially 0. 1/infinity X 1/infinity, right? Is the point of the question to prove it would be impossible to satisfy the conditions of the question?

[David Sklansky]

I said both numbers must be above .6, which happens 16% of the time. And its even money from there.

[diddle]

I said both numbers must be above 1.6 total, which happens 8% of the time. And it's a lock from there.

[madnak]

[ QUOTE ]
oh...I misread the question ;-) I wondered if there was some trick or something.

There are an infinite amount of real numbers between 0 and 1. So the probability of being dealt a hand at all would be essentially 0. 1/infinity X 1/infinity, right? Is the point of the question to prove it would be impossible to satisfy the conditions of the question?

[/ QUOTE ]

The probability of being dealt any specific hand would be infinity. The probability of being dealt a range of hands would not be.

[ALawPoker]

[ QUOTE ]
easy, ill put my calculations here.( it wont make too much sense)

40% of the time we are still alive after the first card.
When its 0,6 we have a 0% chance of winning
when it 1 we have a 40% chance of winning, on average we will have a 20% chane of winning.
There by 0,4 x 0,2 = 0,08

8%.

[/ QUOTE ]

This is how I did it.

Interesting to see the various paths.

[madnak]

[ QUOTE ]
[ QUOTE ]
oh...I misread the question ;-) I wondered if there was some trick or something.

There are an infinite amount of real numbers between 0 and 1. So the probability of being dealt a hand at all would be essentially 0. 1/infinity X 1/infinity, right? Is the point of the question to prove it would be impossible to satisfy the conditions of the question?

[/ QUOTE ]

The probability of being dealt any specific hand would be infinity. The probability of being dealt a range of hands would not be.

[/ QUOTE ]

Er, 0. Not infinity.

[MaxWeiss]

Go to the probability forum and ask. If there is disagreement, then whatever pzhon or a moderator says is true. If they have a different answer, then the universe will collapse and cease to exist.

Duh.